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Using Green's theorem to evaluate

  1. Jan 16, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    Using Green's theorem, evaluate:
    http://s2.ipicture.ru/uploads/20120117/6p57O2HO.jpg

    The attempt at a solution
    [tex]\frac{\partial P}{\partial y}=3x+2y[/tex]
    [tex]\frac{\partial Q}{\partial x}=2y+10x[/tex]
    [tex]\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=7x[/tex]
    To do the integration, i'm using the parametric equations of the circle; x= cosθ and y=sinθ
    [tex]\int\int 7\cos \theta \,.rdrd\theta[/tex]
    The curve C is a circle with centre (1,-2) and radius r=1
    I've drawn the graph in my copybook.

    Description of circle:
    For θ fixed, r varies from r=-2 to r=-3
    θ varies from -∏/2 to -∏/2

    However, i can't get the correct answer as i'm quite sure that i've got the wrong limits. Usually, i deal with circles with centre (0,0) so it's easier to find the limits for r and θ.
     
  2. jcsd
  3. Jan 17, 2012 #2

    sharks

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    Any help?
     
  4. Jan 17, 2012 #3

    LCKurtz

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    Your equations ##x=r\cos\theta,\ y=r\sin\theta## give a circle at the origin, not centered at ##(1,-2)##. First figure out how to fix that (think about translation). Then, standard convention in polar coordinates uses positive r, so it wouldn't go from -2 to -3. And of course if you use the same upper and lower limits for ##\theta##, you will get 0. Think about if ##
    \theta## starts at ##-\frac \pi 2## and increases by ##2\pi##, where will it end?
     
  5. Jan 18, 2012 #4

    sharks

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    Hi LCKurtz

    Here is the Cartesian graph of path C:

    http://s1.ipicture.ru/uploads/20120118/SQzbOw3U.gif

    An attempt at translation of axes:

    Let x = X+h
    Let y=Y+k
    Then, X=rcosθ-h and Y=rsinθ-k

    Here, x=1 and y=-2.
    I'm not sure. Is this correct?

    I'm going to try another method:
    From the general equation for polar coordinates, [itex]x^2+y^2=r^2[/itex], at (1,-2), r=√5
    This would mean that r varies from r=0 to r=√5
    Then, at (1,-2), x=1 and y=-2
    Substituting into x=rcosθ and y=rsinθ
    1=√5cosθ, θ= 1.107
    and -2=√5sinθ, θ=-1.107
    Thus θ varies from θ=-1.107 to θ=1.107
    Again, i'm not sure. Maybe there's a simpler and more accurate method?

    I did some more research and i believe this method is correct:
    The new centre of circle is: [(x-1),(y+2)] and the radius stays same, r=1.
    For [itex]x=r\cos\theta,\ y=r\sin\theta[/itex], these become [itex](x-1)=r\cos\theta,\ (y+2)=r\sin\theta[/itex]
    So, [itex]x=r\cos\theta+1,\ y=r\sin\theta-2[/itex]
    Since, r=1, the parametric equations of the circle become: [itex]x=\cos\theta+1,\ y=\sin\theta-2[/itex]

    Therefore, the integral becomes:
    [tex]\int\int 7(\cos\theta+1) \,.rdrd\theta=\int\int 7\cos\theta+7 \,.rdrd\theta[/tex]
    The description of the circle:
    For θ fixed, r varies from r=0 to r=1.
    θ varies from θ=0 to θ=2∏.

    The integration gives: 7∏, which is the correct answer. So, i assume that i finally found the right method.
     
    Last edited: Jan 18, 2012
  6. Jan 18, 2012 #5

    LCKurtz

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    Yes! Good work, and you learned something in the process. :biggrin:
     
  7. Jan 18, 2012 #6

    sharks

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    Thanks for your guidance, LCKurtz. It's much appreciated.:smile:
     
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